Question

For what value of k, the roots are really and equal
k²x²-2(2k-1)x+1=0
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Answer 1

Answer:

{-2(2k-1)} ^2 - 4(k^2) (1)=0

(-4k+2)^2 - 4k^2=0

16k^2 +4-16k-4k^2 =0

12k^2 - 16k+4=0

4(3k^2 - 4k+1)=0

3k^2 - 4k+1=0

3k^2 - 3k-k+1=0

3k(k-1)-1(k-1)=0

(3k-1)(k-1)=0

K=1/3,K=1

Answer 2

Answer:

if roots are really equal then

b²-4Ac=0

(2(2k-1))²-4(k²)(1)=0

4k-2)²-4k²=0

16k²+4-16k-4k²=0

12k²-16k+4=0

4(3k²-4k+1)=0

3k²-4k+1=0

3k²-3k-1k+1=0

3k(k-1)-1(k-1)=0

values of k are 1 or 1/3

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